2007-08-18 13:36:04 · 5 answers · asked by CPUcate 6 in Science & Mathematics ➔ Mathematics
log(10x) = log[10*10^(.6290/2)] = 1.3145
2007-08-18 13:41:26 · answer #1 · answered by sahsjing 7 · 0⤊ 0⤋
log x^2 = 0.6290
2log x = .6290
log x = .3145
x = 2.063
10x = 20.63
log 10x = log 23.63 = 1.3735
2007-08-18 20:46:08 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
First, solve for X
10^0.6290 =x^2
sqrt(10^0.6290)=x
2.063=x
Substitute for x
log 10(2.063)
log 20.63= 1.314
2007-08-18 20:50:18 · answer #3 · answered by james w 5 · 0⤊ 0⤋
Assuming base 10...
log[10](x^2) = 0.6290
10^(0.629) = x^2
4.256 = x^2
Square root of both sides:
x = ±2.063
log[10](±2.063)
Change of base formula:
= ln(2.063) / ln10 [can't take log of a negative number]
= 0.724 / 2.303
= 0.314
2007-08-18 20:43:39 · answer #4 · answered by whitesox09 7 · 0⤊ 1⤋
It will be half that, or 0.3145. (We assume that the first log is Briggs, as the last one is.)
2007-08-18 20:42:18 · answer #5 · answered by Anonymous · 1⤊ 0⤋